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The range had the wrong bounds, the interval [-1, 1] becomes [0, 2] when adding 1 |
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Differentiating under the sign of integral it is not difficult to check that ''{{prime|φ}}''(''t'') = 0 for all ''t'', so ''φ'' is a constant function, which is a contradiction because ''φ''(0) is the ''n''-dimensional volume of the ball, while ''φ''(1) is zero. The geometric idea is that ''φ''(''t'') is the oriented area of ''g''<sup>''t''</sup>(''B'') (that is, the Lebesgue measure of the image of the ball via ''g''<sup>''t''</sup>, taking into account multiplicity and orientation), and should remain constant (as it is very clear in the one-dimensional case). On the other hand, as the parameter ''t'' passes form 0 to 1 the map ''g''<sup>''t''</sup> transforms continuously from the identity map of the ball, to the retraction ''r'', which is a contradiction since the oriented area of the identity coincides with the volume of the ball, while the oriented area of ''r'' is necessarily 0, as its image is the boundary of the ball, a set of null measure.<ref>{{harvnb|Kulpa|1989}}</ref>
===A proof using the game
A quite different proof given by [[David Gale]] is based on the game of [[Hex (board game)|Hex]]. The basic theorem about Hex is that no game can end in a draw. This is equivalent to the Brouwer fixed-point theorem for dimension 2. By considering ''n''-dimensional versions of Hex, one can prove in general that Brouwer's theorem is equivalent to the [[determinacy]] theorem for Hex.<ref>{{cite journal|author=David Gale |year=1979|title=The Game of Hex and Brouwer Fixed-Point Theorem | journal=The American Mathematical Monthly | volume=86 | pages=818–827|doi=10.2307/2320146|jstor=2320146|issue=10}}</ref>
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